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Q.
If f(x) = (x - 1) (x - 3)(x - 4)(x - 6) + 19 for all real value of x is
Relations and Functions
Solution:
Note: Discarte's Rule of Signs
No equation can have more +ve real roots than it has changes of sign from +ve to -ve, and from -ve to + ve.
And similarly, number of -ve roots of f(x) can not more than the number of changes of sign in f (- x).
Now given
$f (x) = (x - 1) (x - 3) (x - 4) (x - 6) + 19$
$\Rightarrow \, \, f (x) = x^4 - 14x^3 + 67x^2 - 126x + 91$
Here, f(x) has four changes of sign. So the equation can not have more than four positive roots. And f(-x) has no changes of sign, so by Discarte rule, the equation have no relative roots. As
$f(-x) = x^4 + 14x^3 + 67x^2 + 126x + 91$
Thus the equation have four positive roots.